# Simultaneous eigenstates of Hamiltonian and momentum operator

Given the potential barrier, \begin{align} V(x, y) = \left\{ \begin{array}{cc} V_{0} & \hspace{5mm} \text{if 0 \leq x \leq D} \\ 0 & \hspace{5mm} \text{otherwise} \end{array} \right. \end{align}

the Hamiltonian of the system is $$\hat H = -\frac{\hbar^{2}}{2m}\nabla^{2}+V$$

Hence for $$x<0$$, the time-independent wavefunction is: $$\Psi(x) = A\,exp(ikx)+B\, exp(-ikx)$$

This is an eigenvector of $$\hat H$$ with the first term representing the incident wave while the second term represents the reflected wave.

Now for this region $$[\hat H, \hat p]=0$$, so they should have common non-degenerate eigenvectors.

But the above wavefunction is not an eigenvector of $$\hat p$$. What am I thinking wrong here?

• You might want to check this out : physics.stackexchange.com/q/221007. That the commutator is zero does not mean that both operators have the same eigenvectors. What they do posses is a set of simultaneous eigenvectors. – WarreG Jun 12 at 13:32

In your example the difficulty is that you're taking linear combination of eigenstates of $$p$$ but with different eigenvalues, so the resulting combination is no longer an eigenstate of $$p$$, even if the pieces are separately eigenstates.

An alternate example would be the simple case $$[\hat H,\hat L^2]=0$$ and a hydrogen atom state with $$n=2$$ so that $$\ell=0,1$$ can occur. Then $$\{\vert n\ell m\rangle\}$$ are simultaneous eigenvectors of $$\hat H$$ and $$\hat L^2$$ but a combination of these containing different $$\ell$$s will not be a simultaneous eigenstate of both since different $$\ell$$ states have different eigenvalues of $$\hat L^2$$.

• Great example and great answer. – gented Jun 14 at 14:17

If the commutator $$[H,p] = 0$$, it means that there exists a system of eigenvectors which are common to both operators, and that system of eigenvectors spans the Hilbert space of both the operators.

Here the system of simultaneous eigenvectors of $$H$$ and $$p$$ are $$\{e^{ikx}\}$$ for all $$k \in (-\infty, \infty)$$. We call this continuous eigenspectrum. Now if you look at your solution wavefunction, you will see that the solution is a linear superposition of two eigenvectors. This superposition state may not be an eigenvector for the momentum space.

$$\def\hp{{\hat p}} \def\hH{{\hat H}}$$ IMO there is a basic misunderstanding in your arguments, about what operators and eigenfunctions do mean.

You write

Hence for $$x<0$$, the time-independent wavefunction is: $$\Psi(x)=A\,\exp(ikx)+B\,\exp(-ikx)$$

This is an eigenvector of $$\hat H$$

No, this is at least an improper way of saying. It's true that Schr. eqn for stationary states requires a solution of that form for $$x<0$$. But you can't call it an eigenvector. This name is properly used only for a function obeying Schr. eqn on the whole real line.

Analogously, you're not allowed to say

for this region $$[\hH,\hp]=0$$

Operators are defined over a space of functions and no significant conclusion can be drawn from a relationship only holding over a subset of functions domain.

so they should have common non-degenerate eigenvectors

This conclusion would be right if your operators did really commute, all over Hilbert space, not on functions restricted to a subset of real line.

Note that if $$\hH$$ and $$\hp$$ did commute they would have a complete system of common eigenfunctions. This happens for a free particle: here eigenfunctions of $$\hp$$ are $$\exp(ikx)$$ for $$k\in\Bbb R$$. This is a complete set and all are also eigenfunctions of $$\hH$$ although degenerate on $$k$$ sign.

• You're technically correct but your answer would be much improved if you could explain domain issues in connection to the whole real line. – ZeroTheHero Jun 13 at 15:15

The wave-function you have written down is the solution to the "free" wave equation, this is when $$V=cst$$. The correct wave-function is somewhat more complicated to compute and requires an analysis of the boundary conditions at $$x=0,D$$

Another way to see why your Hamiltonian does not commute with the momentum operator is that the potential $$V$$ and thus the Hamiltonian you have is not translation invariant, there are three distinct regions:

$$1.\quad x<0 \\ 2.\quad 0<=x<=D \\ 3.\quad D and the potential is $$V = \Theta(x)-\Theta(x-D)$$ where $$\Theta$$ is the Heaviside step function.